CPhill

I'm making a guess as to what is meant here

x [√(9-4)/3] = 5   multiply both sides by 3

x √5 = 15       divide both sides by √5

x = 15/√5 = (15√5) / 5   =  3√5

sin 240 x sec (-45)-tan(-30) x csc(-315)=

[-√(3)/2]*[√2] - [-1/√(3)]*[√2] =

[-√6/2] + [√2/√3]  =

[-√6/2] + [√6/3]  =

[-3√6]/6 + [2√6]/6  =

-√6 / 6

I assume we have

∫x^(-3/4) dx

So we have

4[16^(1/4) - 1^(1/4)] =

4[2 - 1] =

4 [1]  = 4

(3/4)-(2/5*7/12)

3/4 -[(2/5)*(7/12)]       perform the multiplication in the brackets, first

3/4 - (14/60)         reduce 14/60 = 7/30

3/4 - (7/30)

[30(3) - 4(7)] / 120

62/120 =

31/ 60

22-32-5x3-6=30

(22-3(2-5)x(3))(-6) = 30

(22 -3(3)x(3))(-6) = 30

(22 -9x(3))(-6) = 30

(22-27)(-6) = 30

(-5)(-6) = 30

Whew!!!...that one was pretty tough !!!!

This isn't a perfectly linear relatioship....using a line of "best fit" calculator to model the data in a linear fashion as closely as possible, we have

y = 3.117143x  - 1.676190

y =  3.117143(6)  - 1.676190 ≈ 17.03 on Day 6

y =  3.117143(7)  - 1.676190 ≈ 20.14 on Day 7

There may be other functions that will fit the data more closely - (perhaps a  polynomial?? ).....but I'm not sure......

That's one thing about these identities......there are many approaches, sometimes......

90% x 630,000 = 567,000 voted yes

The linear equation that models this is:

y = (3/7)x + 24    where y is the height of the tree on any given day in inches and x represents the number of days elapsed after day "0."  Note that, on day "0," the tree is 24 inches tall = 2ft.

Here's the graph here: 

P.S. - You can create a table on your own.....just plug a few values of x into the above equation and generate some different y values

1/[csc A - cotA] - 1/sinA = 1/sinA  - 1/[csc A + cot A]     get everything in terms of sine and cosine

1/ [(1/sinA - cosA/sinA)] - 1/sinA = 1/sinA - 1/ [(1/sinA + cosA/sinA)]    simplify this

sinA/[1 - cosA] - 1/sinA = 1/sinA - sinA/[1 + cosA]

Multiply the first term on the LHS by (1 + cos A) in the the numerator and the denominator. The denominator becomes (1-cos^A) = sin^2A.......the same sort of thing is also done to the last term on the RHS....

sinA[1 + cosA]/sin^2A - 1/sinA = 1/sinA - sinA[1-cosA]/sin^2A

Get common denominators - (sin^2A) - on both sides

sinA[1+cosA]/sin^2A -sinA/sin^2A  = sinA/sin^2A - sinA[1 - cosA]/sin^2a

sinA/sin^2a + cosA/sin^A - sinA/sin^2A = sinA/sin^2A - sinA/sin^2A + cosA/sin^2A

cosA/sin^2A    = cosA/sin^2A